考虑层序遍历

如果当前节点的左右子树任一为空,判断同层的下一个节点的左右子树是否都为空即可

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return bool布尔型
     */
    int num = -1;
    public boolean isCompleteTree (TreeNode root) {
        if (root == null) {
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            while (size-- > 0) {
                TreeNode node = queue.poll();
                if (node.left == null && node.right != null) {
                    return false;
                }
                if (num > -1 && (node.left != null || node.right != null)) {
                    return false;
                }
                if (node.left == null || node.right == null) {
                    num = 0;
                }
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }
        return true;
    }
}